Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $n \neq 0$. $y = \dfrac{-8n + 64}{2n - 6} \times \dfrac{n^2 - 5n + 6}{9n - 18} $
Answer: First factor the quadratic. $y = \dfrac{-8n + 64}{2n - 6} \times \dfrac{(n - 3)(n - 2)}{9n - 18} $ Then factor out any other terms. $y = \dfrac{-8(n - 8)}{2(n - 3)} \times \dfrac{(n - 3)(n - 2)}{9(n - 2)} $ Then multiply the two numerators and multiply the two denominators. $y = \dfrac{ -8(n - 8) \times (n - 3)(n - 2) } { 2(n - 3) \times 9(n - 2) } $ $y = \dfrac{ -8(n - 8)(n - 3)(n - 2)}{ 18(n - 3)(n - 2)} $ Notice that $(n - 2)$ and $(n - 3)$ appear in both the numerator and denominator so we can cancel them. $y = \dfrac{ -8(n - 8)\cancel{(n - 3)}(n - 2)}{ 18\cancel{(n - 3)}(n - 2)} $ We are dividing by $n - 3$ , so $n - 3 \neq 0$ Therefore, $n \neq 3$ $y = \dfrac{ -8(n - 8)\cancel{(n - 3)}\cancel{(n - 2)}}{ 18\cancel{(n - 3)}\cancel{(n - 2)}} $ We are dividing by $n - 2$ , so $n - 2 \neq 0$ Therefore, $n \neq 2$ $y = \dfrac{-8(n - 8)}{18} $ $y = \dfrac{-4(n - 8)}{9} ; \space n \neq 3 ; \space n \neq 2 $